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Here,first we examine whether the form under the integral is exact,

Take F=2xyz^{2}i+(x^{2}z^{2}+z\cos yz)j+(2x^{2}yz+y\cos yz)k\,

Therefore,{\mathrm  {curl}}F={\begin{vmatrix}i&j&k\\{\frac  {\partial }{\partial x}}&{\frac  {\partial }{\partial y}}&{\frac  {\partial }{\partial z}}\\2xyz^{2}&(x^{2}z^{2}+z\cos yz)&(2x^{2}yz+y\cos yz)\end{vmatrix}}\,

=(2x^{2}z+\cos yz-yz\sin yz-2x^{2}z-\cos yz+yz\sin yz)i-(4xyz-4xyz)j+(2xz^{2}-2xz^{2})k=0\,

showing that the given integral is independent of the path in space and there exists a function \phi (x,y,z)\, such that F=\nabla \phi \, so that

2xyz^{2}i+(x^{2}z^{2}+z\cos yz)j+(2x^{2}yz+y\cos yz)k={\frac  {\partial \phi }{\partial x}}+{\frac  {\partial \phi }{\partial y}}j+{\frac  {\partial \phi }{\partial z}}k\,

Therefore,{\frac  {\partial \phi }{\partial x}}=2xyz^{2}\, implies \phi =x^{2}yz^{2}+f_{1}(y,z)\, --(1)

{\frac  {\partial \phi }{\partial y}}=x^{2}z^{2}+z\cos yz\, implies \phi =x^{2}z^{2}y+\sin yz+f_{2}(x,z)\, --(2)

{\frac  {\partial \phi }{\partial z}}=2x^{2}yz+y\cos z\, implies \phi =x^{2}yz^{2}+\sin yz+f_{3}(x,y)\, --(3)

Now (1),(2),(3) each represent phi. These agree if we choose f_{1}(y,z)=\sin yz,f_{2}(z,x)=0,f_{3}(x,y)=0\,

Therefore \phi =x^{2}yz^{2}+\sin yz\,

Hence the given integral =\int _{C}F\cdot dr=\int _{C}\nabla \phi \cdot dr=\int d\phi \,

=[x^{2}yz^{2}+\sin yz]_{{0,0,1}}^{{1,{\frac  {\pi }{4}},2}}=\pi +\sin {\frac  {\pi }{2}}=\pi +1\,

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